solve: tanx+tan2x+√3tanx tan2x=√3 (0<=x<=360)

tanx+tan2x+√3tanx tan2x=√3

tanx+tan2x = √3 - √3tanxtan2x
tanx+tan2x = √3(1-tanx tan2x)
(tanx+tan2x)/(1 - tanx tan2x) = √3
tan3x = √3
...

What a splendid observation !

Ignore my gibberish, it must contain an error somewhere anyhow.

To solve the equation tan(x) + tan(2x) + √3tan(x)tan(2x) = √3 in the range 0 <= x <= 360 degrees, we can use some trigonometric identities and algebraic manipulation. Here's how to do it step by step:

Step 1: Simplify the expression using the double-angle identity for tangent:
tan(2x) = 2tan(x) / (1 - tan^2(x))

The equation now becomes:
tan(x) + (2tan(x) / (1 - tan^2(x))) + √3tan(x)(2tan(x) / (1 - tan^2(x))) = √3

Step 2: Multiply both sides of the equation by (1 - tan^2(x)) to eliminate the denominator:
[(1 - tan^2(x))tan(x)] + 2tan(x) + 2√3tan^2(x) = √3(1 - tan^2(x))

Simplifying further:
tan(x) - tan^3(x) + 2tan(x) + 2√3tan^2(x) = √3 - √3tan^2(x)

Step 3: Rearrange the terms and simplify by combining like terms:
√3tan^2(x) - √3tan^3(x) - 2tan^2(x) + 2tan(x) + √3 = 0

Step 4: Factor out common factors:
tan^2(x)(√3 - 2) - tan(x)(√3tan^2(x) - 2) + √3 = 0

Step 5: Set each factor equal to zero and solve for x:
tan^2(x) = 0 --> x = 0, 180

√3tan^2(x) - 2 = 0 --> tan^2(x) = 2/√3 --> x = 30, 210

√3tan^2(x) - 2tan(x) = 0
We can factor out tan(x):
tan(x)(√3tan(x) - 2) = 0

Therefore, either tan(x) = 0 or (√3tan(x) - 2) = 0

If tan(x) = 0, then x = 0, 180

If √3tan(x) - 2 = 0, then tan(x) = 2/√3 --> x = 30, 210

So, the solutions for the equation tan(x) + tan(2x) + √3tan(x)tan(2x) = √3 in the range 0 <= x <= 360 are:
x = 0, 30, 180, 210

tanx - ?3 + tan2x + ?3tanx tan2x = 0

tanx - ?3 + tan 2x(1 + ?3tanx) = 0

recall tan 2x = 2tanx/(1 - tan^2 x)

tanx - ?3 + 2tanx/(1 - tan^2 x)(1 + ?3tanx) = 0
times (1 - tan^2 x)

tanx - tan^3 x - ?3 + ?3tan^2 x + 2tanx(1 + ?3tanx) = 0
let tanx = y for easier typing
y - y^3 - ?3 - ?3y^2 + 2y(1 + ?3y) = 0
y - y^3 - ?3 - ?3y^2 + 2y + 2?3y^2 = 0
y^3 - ?3y^2 - 3y - ?3 = 0

not easy to solve a cubic, so I sent it to
Wolfram:
http://www.wolframalpha.com/input/?i=y+-+y%5E3+-+%E2%88%9A3+-+(%E2%88%9A3)y%5E2+%2B+2y+%2B+(2%E2%88%9A3)y%5E2+%3D+0

See if that helps